Let w and l be the dimensions (width and length, respectively) of the coop.
We know that the length of the coop is 4 feet less than twice the width, which means that
[tex]l=2w-4[/tex]
Also, the area is 510, but the area is the product of the dimensions, so we have
[tex]lw=510[/tex]
Plug the expression for l in the formula for the area:
[tex]lw=(2w-4)w=510 \iff 2w^2-4w-510=0[/tex]
We can divide the whole expression by 2 and solve it with the quadratic formula:
[tex]w^2-2w-255=0 \iff w=\dfrac{2\pm\sqrt{4+1020}}{2}=\dfrac{2\pm 32}{2}=1\pm 16[/tex]
So, the two solutions are
[tex]w_1=1-16=-15,\quad w_2=1+16=17[/tex]
The negative solution makes no sense (we can't have negative lengths), so the width must be 17.
We conclude that the length is
[tex]l=2w-4=2\cdot 17-4=34-4=30[/tex]
